Problem: Let $y=5^{(6x^2-x+2)}$. Find $\dfrac{dy}{dx}$. Choose 1 answer: Choose 1 answer: (Choice A) A $\ln(5)\cdot 5^{(12x-1)}$ (Choice B) B $\ln(5)\cdot 5^{(6x^2-x+2)}\cdot (12x-1)$ (Choice C) C $\ln(5)\cdot 5^{(6x^2-x+2)}$ (Choice D) D $ 5^{(6x^2-x+2)}\cdot (12x-1)$
$5^{(6x^2-x+2)}$ is an exponential function, but its argument isn't simply $x$. Therefore, it's a composite function. In other words, suppose $u(x)=6x^2-x+2$, then $y=5^{u(x)}$. $\dfrac{dy}{dx}$ can be found using the following identity: $\dfrac{d}{dx}\left[5^{u(x)}\right]=\ln(5)\cdot 5^{u(x)}\cdot u'(x)$ [Why is this identity true?] Let's differentiate! $\begin{aligned} &\phantom{=}\dfrac{dy}{dx} \\\\ &=\dfrac{d}{dx}5^{(6x^2-x+2)} \\\\ &=\dfrac{d}{dx}5^{u(x)}&&\gray{\text{Let }u(x)=6x^2-x+2} \\\\ &=\ln(5)\cdot 5^{u(x)}\cdot u'(x) \\\\ &=\ln(5)\cdot5^{(6x^2-x+2)}\cdot (12x-1)&&\gray{\text{Substitute }u(x)\text{ back}} \end{aligned}$ In conclusion, $\dfrac{dy}{dx}=\ln(5)\cdot5^{(6x^2-x+2)}\cdot (12x-1)$